College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 59


$x \approx 2.60$

Work Step by Step

Take the natural log of both sides to obtain: $\ln{(2^{3x-5})} = \ln{7}$ Use the rule $\ln{(a^n)} = n\cdot\ln{a}$ to obtain: $(3x-5)\ln{2} = \ln{7}$ Divide $\ln{2}$ to both sides of the equation to obtain: $\dfrac{(3x-5)\ln2}{\ln2} = \dfrac{\ln7}{\ln2} \\3x-5=\dfrac{\ln7}{\ln2}$ Add $5$ on both sides of the equation to obtain: $3x=5+\dfrac{\ln7}{\ln2}$ Divide $3$ on both of the equation to obtain: $x = \dfrac{5+\frac{\ln7}{\ln2}}{3}$ Use a calculator to obtain: $x \approx 2.60$
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